**NCERT Exemplar Class 10 Maths Chapter 9 Circles** are part of NCERT Exemplar Class 10 Maths. Here we have given NCERT Exemplar Class 10 Maths Chapter 9 Circles.

## NCERT Exemplar Class 10 Maths Chapter 9 Circles

**Exercise 9.1**

Choose the correct answer from the given four options:

Question 1

If radii of two concentric circles are 4 cm and 5 cm, then the length of each chord of one circle which is tangent to the other circle is

(A) 3 cm

(B) 6 cm

(C) 9 cm

(D) 1 cm

Solution:

(B) Let O be the centre of two concentric circles C_{1} and C_{2}, whose radii are r_{1} = 4 cm and r_{2} = 5 cm.

Now, we draw a chord AC of circle C_{2}, which touches the circle Q at B.

Also, join OB, which is perpendicular to AC.

[∵ Tangent at any point of circle is perpendicular to radius through the point of contact]

Now, in right angled ∆OBC,

OC^{2} = BC^{2} + BO^{2} [By Pythagoras theorem]

⇒ 5^{2} = BC^{2} + 4^{2}

⇒ BC^{2} = 25 – 16 = 9

⇒ BC = 3 cm

∴ Length of chord AC = 2BC = 2 x 3 = 6 cm

Question 2

In figure, if ∠AOB = 125°, then ∠COD is equal to

(A) 62.5°

(B) 45°

(C) 35°

(D) 55°

Solution:

(D) We know that the opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.

⇒ ∠AOB + ∠COD = 180°

⇒ ∠COD = 180° – ∠AOB

= 180° – 125° = 55°

Question 3

In figure, AB is a chord of the circle and AOC is its diameter such that ∠ACB = 50°. If AT is the tangent to the circle at the point A, then ∠BAT is equal to

(A) 65°

(C) 50°

(B) 60°

(D) 40°

Solution:

(C) In figure, AOC is a diameter of the circle. We know that, diameter subtends an angle 90° at the circle.

So, ∠ABC = 90°

In ∆ACB,

∠A + ∠B + ∠C = 180° [Angle sum property]

⇒ ∠A + 90° + 50° = 180°

⇒ ∠A + 140° = 180°

⇒ ∠A = 180° – 140° = 40°

⇒ ∠A or ∠OAB = 40°

Since, AT is the tangent to the circle at point A. Therefore, OA is perpendicular to AT.

∠OAT = 90°

⇒ ∠OAB + ∠BAT =90°

On putting ∠OAB = 40°, we get

⇒ ∠BAT =90° -40° = 50°

Hence, the value of ∠BAT is 50°.

Question 4

From a point P which is at a distance of 13 cm from the centre O of a circle of radius 5 cm, the pair of tangents PQ and PR to the circle are drawn. Then the area of the quadrilateral PQOR is

(A) 60 cm^{2}

(B) 65 cm^{2}

(C) 30 cm^{2}

(D) 32.5 cm^{2}

Solution:

(A) Firstly, draw a circle of radius 5 cm with centre O. P is a point at a distance of 13 cm from O. A pair of tangents PQ and PR are drawn.

Thus, quadrilateral PQOR is formed.

∵ OQ ⊥ QP [since, QP is a tangent line]

In right angled ∆PQO,

OP^{2} = OQ^{2} + QP^{2}

⇒ 13^{2} = 5^{2} + QP^{2}

⇒ QP^{2} = 169 – 25 = 144

⇒ QP = 12 cm

Now, area of ∆OQP = \(\frac { 1 }{ 2 }\) x QP x QO

= \(\frac { 1 }{ 2 }\) x 12 x 5 = 30 cm^{2}

∴ Area of quadrilateral PQOR = 2 x ar ∆OQP

= 2 x 30 = 60 cm^{2}

Question 5

At one end A of a diameter AB of a circle of radius 5 cm, tangent XAY is drawn to the circle. The length of the chord CD parallel to XT and at a distance 8 cm from A is

(A) 4 cm

(B) 5 cm

(C) 6 cm

(D) 8 cm

Solution:

(D) First, draw a circle of radius 5 cm with centre O. A tangent XY is drawn at point A.

A chord CD is drawn which is parallel to XY and at a distance of 8 cm from A.

Now, ∠OAY = 90°

[ ∵ Tangent at any point of circle is perpendicular to the radius through the point of contact]

∠OAY +∠OED= 180°

[ ∵ Sum of cointerior angles is 180°]

⇒ ∠OED = 180° – 90° = 90°

Also, AE = 8 cm Join OC.

OC = 5 cm [Radius of circle]

OE = AE-OA = 8 – 5 = 3 cm

Now, in right angled ∆OEC,

OC^{2} = OE^{2} + EC^{2} [By Pythagoras theorem]

⇒ EC^{2} = OC^{2} – OE^{2}

⇒ EC^{2} = 5^{2} – 3^{2}

⇒ EC^{2} = 25 – 9 = 16

⇒ EC = 4 cm

Since, perpendicular from centre to the chord bisects the chord.

∴ CE = ED

⇒ CD = 2 x EC

⇒ CD = 2 x 4

⇒ CD = 8 cm

Question 6

In figure, ATis a tangent to the circle with centre O such that 07 = 4 cm and ∠OTA = 30°. Then AT is equal to

(A) 4cm

(B) 2cm

(C) 2√3cm

(D) 4√3cm

Solution:

(C) Join OA.

We know that, the tangent at any point of a circle is perpendicular to the radius through the point of contact.

Question 7

In figure, if O is the centre of a circle, PQ is a chord and the tangent PR at P makes an angle of 500 with PQ, then ∠POQ is equal to

(A) 100°

(B) 80°

(C) 90°

(D) 75°

Solution:

(A) Given, ∠QPR = 50°

We know that, the tangent at any point of a circle is perpendicular to the radius through the point of contact.

∴ ∠OPR = 90°

⇒ ∠OPQ + ∠QPR = 90°

⇒ ∠OPQ = 90° – 50° = 40° [ ∵ ∠QPR = 50°]

Now, OP = OQ = radius of circle

∴ ∠OQP = ∠OPQ = 40°

[Angles opposite to equal sides are equal]

In ∆OPQ,

∠O + ∠OPQ + ∠Q = 180° [Angle sum property]

⇒ ∠POQ = 180° – (40 + 40°)

= 180° – 80°

[∵ ∠OPQ = 40° = ∠Q]

⇒ ∠POQ = 100°

Question 8

In figure, if PA and PB are tangents to the circle with centre O such that ∠APB = 50°, then ∠OAB is equal to

(A) 25°

(B) 30°

(C) 40°

(D) 50°

Solution:

(A) Given, PA and PB are tangent lines.

∴ PA = PB [∵ Length of tangents drawn from an external point to a circle is equal]

∠PBA = ∠PAB = θ (say)

In ∆PAB,

∠P + ∠A + ∠B = 180° [Angle sum property]

⇒ 50° + θ + θ= 180°

⇒ 20 = 180° – 50° = 130°

⇒ θ = 65°

Also, OA ⊥ PA

[∵ Tangent at any point of a circle is perpendicular to the radius through the point of contact]

∴ ∠PAO = 90°

⇒ ∠PAB + ∠BAO = 90°

⇒ 65° + ∠BAO = 90°

⇒ ∠BAO = 90° – 65° = 25°

⇒ ∠OAB = 25°

Question 9

If two tangents inclined at an angle 60° are drawn to a circle of radius 3 cm, then length of each tangent is equal to

Solution:

(D) Let P be an external point and a pair of tangents is drawn from point P such that the angle between two tangents is 60°.

Join OA and OP.

Also, OP is a bisector line of ∠APC.

∴ ∠APO = ∠CPO = 30°

Also, OA ⊥ AP

[Tangent at any point of a circle is perpendicular to the radius through the point of contact.]

∴∠OAP = 90°

In right angled ∆OAP,

Hence, the length of each tangent is 3√3 cm.

Question 10

In figure, if PQR is the tangent to a circle at Q whose centre is O, AB is a chord parallel to PR and ∠BQR = 70°, then ∠AQB is equal to

(A) 20°

(B) 40°

(C) 35°

(D) 45°

Solution:

(B) Given, AB ॥ PR

∴ ∠ABQ = ∠BQR = 70° [Alternate angles]

Also, QD is perpendicular to AB and QD bisects AB.

In ∆QDA and ∆QDB,

∠QDA = ∠QDB [90° each]

AD = BD

QD = QD [Common side]

∆ADQ ≅ ∆BDQ [by SAS congruency]

⇒ ∠QAD = ∠QBD [CPCT] …(i)

But ∠QBD = ∠ABQ = 70°

⇒ ∠QAD = 70° [From (i)]

Now, in ∆ABQ, ∠A + ∠B + ∠AQB = 180°. [Angle sum property]

⇒ ∠AQB = 180° – (70° + 70°) = 40°

**Exercise 9.2**

Write ‘True’ or ‘False’ and justify your answer in each of the following:

Question 1

If a chord AB subtends an angle of 60° at the centre of a circle, then angle between the tangents at A and B is also 60°.

Solution:

False

Since a chord AB subtends an angle of 60° at the centre of a circle.

⇒ ∠AOB = 60°

As OA = OB = Radius of the circle

∠OAB = ∠OBA = 60°

[∵ Angles opposite to equal sides are equal]

The tangent at points A and B is drawn, which intersect at C.

We know, OA ⊥ AC and OB ⊥ BC.

∴ ∠OAC = 90° and ∠OBC = 90°

⇒ ∠OAB + ∠B AC = 90° and ∠OBA + ∠ABC = 90°

⇒ ∠BAC = 90° – 60° = 30° and ∠ABC = 90° – 60° = 30°

In ∆ABC,

∠BAC + ∠CBA + ∠ACB = 180°

[Angle sum property]

⇒ ∠ACB = 180° – (30° + 30°) = 120°

Question 2

The length of tangent from an external point on a circle is always greater than the radius of the circle.

Solution:

False

Because the length of tangent from an external point P on a circle may or may not be greater than the radius of the circle.

Question 3

The length of tangent from an external point P on a circle with centre O is always less than OP.

Solution:

True

PT is a tangent drawn from an external point P. Join OT

∵ OT ⊥ PT

So, OTP is a right angled triangle formed.

In right angled triangle, hypotenuse is always greater than any of the two sides of the triangle.

∴ OP > PT or PT< OP

Question 4

The angle between two tangents to a circle may be 0°.

Solution:

True

This may be possible only when both tangent lines coincide or are parallel to each other.

Question 5

If angle between two tangents drawn from a point P to a circle of radius a and centre O is 90°, then OP = a√2.

Solution:

True

From point P, two tangents are drawn.

Given, OT = a

Also, line OP bisects the ∠RPT

∴ ∠TPO = ∠RPO = 45°

Also, OT ⊥ TP ⇒ ∠OTP = 90°

In right angled ∆OTP,

Question 6

If angle between two tangents drawn from a point P to a circle of radius a and centre 0 is 60°, then OP = a√3.

Solution:

False

From point P, two tangents are drawn.

Given, OT= a

Also, line OP bisects the ∠RPT.

∴∠TPO = ∠RPO = 30°

Also, OT ⊥ PT

⇒ ∠OTP = 90°

In right angled ∆OTP,

sin 30° = \(\frac { OT }{ OP }\) ⇒ \(\frac { 1 }{ 2 }\) = \(\frac { a }{ OP }\)

⇒ OP = 2a

Question 7

The tangent to the circumcircle of an isosceles triangle ABC at A, in which AB = AC, is parallel to BC.

Solution:

True

Let EAF be tangent to the circumcircle of ∆ABC.

To prove: EAF ॥ BC

We have, ∠EAB = ∠ACB …(i)

[Angle between tangent and chord is equal to angle made by chord in the alternate segment]

Here, AB = AC

⇒ ∠ABC = ∠ACB …(ii)

From equation (i) and (ii), we get

∠EAB = ∠ABC

∵ Alternate angles are equal.

⇒ EAF ॥ BC

Question 8

If a number of circles touch a given line segment PQ at a point A, then their centres lie on the perpendicular bisector of PQ.

Solution:

False

Given that PQ is any line segment and S_{1}, S_{2}, S_{3}, S_{4}, …….. circles touch the line segment PQ at a point A. Let the centres of the circles S_{1}, S_{2}, S_{3}, S_{4}, ……….. be C_{1}, C_{2}, C_{3}, C_{4}, ……… respectively.

To prove: Centres of the circles lie on the perpendicular bisector of PQ.

Joining each centre of the circles to the point A on the line segment PQ by line segment i.e., C_{1}A, C_{2}A, C_{3}A, C_{4}A,… and so on.

We know that, if we draw a line from the centre of a circle to its tangent line, then the line is always perpendicular to the tangent line. But it does not bisect the line segment PQ.

Question 9

If a number of circles pass through the end points P and Q of a line segment PQ, then their centres lie on the perpendicular bisector of PQ.

Solution:

True

We draw two circles with centre C_{1} and C_{2} passing through the end points P and Q of a line segment PQ. We know that the perpendi-cular bisector of a chord of circle always passes through the centre of the circle.

Thus, perpendicular bisector of PQ passes through C_{1} and C_{2} . Similarly, all the circle passing through the end points of line segment PQ, will have their centres on the perpendicular bisector of PQ.

Question 10

AB is a diameter of a circle and AC is its chord such that ∠BAC = 30°. If the tangent at C intersects AB extended at D, then BC = BD.

Solution:

True

To prove: BC = BD

Join BC and OC.

Given, ∠BAC = 30°

⇒ ∠BCD = 30°

[Angle between tangent and chord is equal to angle made by chord in the alternate segment]

∵ OC ⊥ CD and OA = OC = radius

⇒ ∠OAC = ∠OCA = 30°

∠ACD = ∠ACO + ∠OCD = 30° + 90° = 120°

In ∆ACD,

∠DAC + ∠ACD + ∠CDA = 180° [Angle sum property]

⇒ 30° + 120° + ∠CDA = 180°

⇒ ∠CDA = 180° – (30° + 120°) = 30°

∠CDA = ∠BCD

⇒ BC = BD

[∵ Sides opposite to equal angles are equal]

**Exercise 9.3**

Question 1

Out of the two concentric circles, the radius of the outer circle is 5 cm and the chord AC of length 8 cm is a tangent to the inner circle. Find the radius of the inner circle.

Solution:

Let C_{1} and C_{2} be the two circles having same centre O. AC is a chord which touches C_{1} at point D.

Join OD.

Also, OD ⊥ AC

∴ AD = DC = 4 cm

[Perpendicular line OD bisects the chord]

In right angled ∆AOD,

OA^{2} = AD^{2} + OD^{2} [By Pythagoras theorem]

⇒ OD^{2} = 52 – 42

⇒ OD^{2} = 25 – 16 = 9

⇒ OD = 3 cm

∴ Radius of the inner circle is OD = 3 cm

Question 2

Two tangents PQ and PR are drawn from an external point to a circle with centre O. Prove that QORP is a cyclic quadrilateral.

Solution:

To prove: QORP is a cyclic quadrilateral.

Since, PR and PQ are tangents.

∴ OR ∠ PR and OQ ∠ PQ

⇒ ∠ORP = ∠OQP = 90°

Hence, ∠ORP + ∠OQP = 180°

Also, in quadrilateral QORP,

∠ORP + ∠OQP + ∠ROQ + ∠QPR = 360°

⇒ 180° + ∠ROQ + ∠QPR = 360°

⇒ ∠ROQ + ∠QPR = 180°

∵ If sum of opposite angles in quadrilateral is 180°, then the quadrilateral is cyclic.

⇒ QORP is cyclic quadrilateral.

Question 3

If from an external point B of a circle with centre O, two tangents BC and BD are drawn such that ∠DBC = 120°, prove that BC + BD = BO, i.e., BO = 2BC.

Solution:

To prove: BO = 2BC

Given, ∠DBC = 120°

Join OC, OD and BO.

Since, BC and BD are tangents.

∴ OC ⊥ BC and OD ⊥ BD

We know, OB is the angle bisector of ∠DBC.

∴∠OBC = ∠DBO = 60°

In right angled ∆OBC,

⇒ OB = 2 BC

Also, BC = BD

[Tangent drawn from external point to circle are equal]

OB = BC + BC

⇒ OB = BC + BD

Question 4

Prove that the centre of a circle touching two intersecting lines lies on the angle bisector of angle formed by tangents..

Solution:

Given, two tangents PQ and PR are drawn from an external points P to a circle with centre O.

To prove: Centre of a circle touching two intersecting lines lies on the angle bisector of angle formed by tangents.

Construction: Join OR and OQ.

In ∆POR and ∆POQ,

∠PRO = ∠PQO = 90°

[tangent at any point of a circle is perpendicular to the radius through the point of contact]

OR = OQ [Radii of same circle]

OP = OP [Common side]

∆ PRO ≅ ∆ PQO [RHS criterion]

Hence, ∠RPO = ∠QPO [By CPCT]

Thus, O lies on angle bisector of ∠RPQ.

Question 5

In figure, AB and CD are common tangents to two circles of unequal radii.

Prove that AB = CD.

Solution:

Given, AB and CD are common tangents to two circles of unequal radii.

To prove : AB = CD

Construction: Extend AB and CD, to intersect at P.

PA = PC

[∵ Length of tangents drawn from an external point to a circle is equal]

Also, PB = PD

∴PA – PB = PC – PD

⇒ AB = CD

Question 6

In Question 5 above, if radii of the two circles are equal, prove that AB = CD.

Solution:

Join OO’

Since, OA = O’B [Given]

Also, ∠OAB = ∠O’BA = 90°

[Tangent at any point of a circle is perpendicular to the radius at the point of contact]

Since, perpendicular distance between two straight lines at two different points is same.

⇒ AB is parallel to OO’

Similarly, CD is parallel to OO’

⇒AB ॥ CD

Also, ∠OAB = ∠OCD = ∠O’BA = ∠O’DC = 90°

⇒ ABCD is a rectangle.

Hence, AB = CD.

Question 7

In figure, common tangents AB and CD to two circles intersect at E.

Prove that AB = CD.

Solution:

Given, common tangents AB and CD to two circles intersecting at E.

To prove: AB = CD

∵ The length of tangents drawn from an external point to a circle is equal

∴ EB = ED and EA = EC

On adding, we get

EA + EB = EC + ED

⇒ AB = CD

Question 8

A chord PQ of a circle is parallel to the tangent drawn at a point R of the circle. Prove that R bisects the arc PRQ.

Solution:

Given, chord PQ is parallel to tangent at R.

To prove : R bisects the arc PRQ.

∠1 = ∠2 [Alternate interior angles]

∠1 = ∠3 [Angle between tangent and chord is equal to angle made by chord in alternate segment]

⇒ ∠2 = ∠3

⇒ PR = QR [Sides opposite to equal angles are equal]

⇒ arc PR = arc QR

So, R bisects PQ.

Question 9

Prove that the tangents drawn at the ends of a chord of a circle make equal angles with the chord.

Solution:

To prove : ∠1 = ∠2

Let RQ be a chord of the circle. Tangents are drawn at the points R and Q.

Let P be another point on the circle, then join PQ and PR.

Since, at point Q, there is a tangent.

∠2 = ∠P [Angles in alternate

segments are equal] Since, at point R, there is a tangent.

∴ ∠1 = ∠P [Angles in alternate segments are equal]

∴ ∠1 = ∠2 = ∠P

⇒ ∠1 = ∠2

Question 10

Prove that a diameter AB of a circle bisects all those chords which are parallel to the tangent at the point A.

Solution:

Given, AB is a diameter of the circle.

A tangent is drawn at point A.

Draw a chord CD parallel to tangent MAN.

So, CD is a chord of the circle and OA is radius of the circle.

∴ ∠MAO = 90°

[tangent at any point of a circle is perpendicular to the radius through the point of contact]

⇒ ∠CEO = ∠MAO [Corresponding angles]

∴ ∠CEO = 90°

Thus, OE bisects CD,

[Perpendicular from centre of circle to chord bisects the chord]

Similarly, diameter AB bisects all chords which are parallel to the tangent at the point A.

**Exercise 9.4**

Question 1

If a hexagon ABCDEF circumscribe a circle, prove that AB + CD + EF = BC + DE + FA.

Solution:

Given, a hexagon ABCDEF circumscribes a circle.

To prove : AB + CD + EF = BC + DE + FA

∵ AQ = AP

QB = BR

CS = CR

DS = DT

EU = ET

UF = FP

[Tangents drawn from an external point to a circle are equal]

AB + CD + EF = (AQ + QB) + (CS + SD) + (EU + UF) = (AP + BR) + (CR + DT) + (ET + FP)

= (AP + FP) + (BR + CR) + (DT + ET)

= AF + BC + DE

⇒ AB + CD + EF = AF + BC + DE

Question 2

Let s denote the semi-perimeter of a triangle ABC in which BC = a, CA = b, AB = c. If a circle touches the sides BC, CA, AB at D, E, F respectively, prove that BD = s – b.

Solution:

Given, BC = a, CA = b and AB = c

Since, tangents drawn from an external point to the circle are equal in length.

∴ BD = BF = x (say)

DC = CE = y (say)

and AE = AF = 2 (say)

Now, BC + CA + AB = a + b + c

⇒ (BD + DC) + (CE + EA) + (AF + FB) = a + b + c

⇒ (x + y) + (y + z) + (z + x) = a + b + c

⇒ 2(x + y + z) = 2s

[ ∵ 2s = a + b + c = perimeter of ∆ABC]

⇒ s = x + y + z

⇒ x = s – (y + z)

⇒ BD = s – b [∵ b = AE + EC = z + y]

Question 3

From an external point P, two tangents, PA and PB are drawn to a circle with centre O. At one point E on the circle tangent is drawn which intersects PA and PB at C and D, respectively. If PA = 10 cm, find the perimeter of the triangle PCD.

Solution:

∵ Tangents from an external point to a circle are equal in length.

∴ CE = CA, DE = DB and PA = PB.

Perimeter of ∆PCD = PC + CD + PD

= PC + CE + ED + PD

= PC + CA + DB + PD

= PA + PB

= 2PA = 2 x 10 = 20 cm

Question 4

If AB is a chord of a circle with centre O, AOC is a diameter and AT is the tangent at A as shown in figure. Prove that ∠BAT = ∠ACB

Solution:

Since, AC is a diameter line, so angle in semi-circle formed is 90°.

i.e., ∠ABC = 90°

In ∆ABC,

∠CAB + ∠ABC + ∠BCA = 180° [Angle sum property]

⇒ ∠CAB + ∠BCA = 180°-90° = 90° ,..(i)

Since, diameter of a circle is perpendicular to the tangent.

i.e., CA ⊥ AT

∴ ∠CAT = 90°

⇒ ∠CAB + ∠BAT =90° …(ii)

From equation (i) and (ii),

∠CAB + ∠ACB = ∠CAB + BAT

⇒ ∠ACB = ∠BAT

Question 5

Two circles with centres O and O’ of radii 3 cm and 4 cm, respectively intersect at two points P and Q such that OP and O’P are tangents to the two circles. Find the length of the common chord PQ.

Solution:

Here, two circles are of radii OP = 3 cm and O’P = 4 cm.

These two circles intersect at P and Q.

Here, OP and O’P are two tangents drawn at point P.

∠OPO’ = 90° [Tangent at any point of circle is perpendicular to radius through the point of contact]

Join OO’ and PQ such that OO’ and PQ intersect at point N.

In right angled ∆OPO’,

(OO’)^{2} = (OP)^{2} + (PO’)^{2} [By Pythagoras theorem]

⇒ (OO’)^{2} = (3)^{2} + (4)^{2} = 25

⇒ OO’ = 5 cm

Also, PN ⊥ OO’

Let ON = x, then NO’ = 5 – x

In right angled ∆ONP,

(OP)^{2} = (ON)^{2} + (NP)^{2} [By Pythagoras theorem]

⇒ (NP)^{2} = 3^{2} – x^{2} = 9 – x^{2} …(i)

and in right angled ∆PNO’,

(PO’)^{2} = (PN)^{2} + (NO’)^{2} [By Pythagoras theorem]

⇒ (4)^{2} = (PN)^{2} + (5 – x)^{2}

⇒ (PN)^{2} = 16- (5 -x)^{2} …(h)

From equation (i) and (ii),

9 -x^{2} = 16 – (5 -x)^{2}

⇒ 7 + x^{2} – (5 – x)^{2} = 0

⇒ 7 + x^{2} -(25 + x^{2} – 10x) = 0

⇒ 10x = 18

∴ x = 1.8

Again, in right angled ∆OPN,

OP^{2} = (ON)^{2} + (NP)^{2} [By Pythagoras theorem]

⇒ 3^{2}= (1.8)^{2} + (NP)^{2}

⇒ (NP)^{2} = 9 – 3.24 = 5.76

⇒ NP = 2.4

∴ Length of common chord,

PQ = 2PN = 2 x 2.4 = 4.8 cm

Question 6

In a right triangle ABC in which ∠B = 90°, a circle is drawn with AB as diameter intersecting the hypotenuse AC at P. Prove that the tangent to the circle at P bisects BC.

Solution:

Let O be the centre of the given circle. Suppose, the tangent at P meets BC at Q. Join BP.

To prove: BQ = QC

∠ABC = 90° [Given]

In ∆ABC, ∠1 + ∠5 = 90° [Angle sum property, ∠ABC = 90°]

∠3 = ∠1 [Angle between tangent and the chord equals angle made by the chord in alternate segment]

⇒ ∠3 + ∠5 = 90° …(i)

Also, ∠APB = 90° [Angle in semi-circle]

∠APB + ∠BPC = 180°

∴ 90° + ∠3 + ∠4 = 180°

⇒ ∠3 + ∠4 = 90° …(ii)

From equations (i) and (ii), we get ∠3 + ∠5 = ∠3 + ∠4

⇒ ∠5 = ∠4

⇒ PQ = QC [Sides opposite to equal angles are equal]

Also, QP = QB

[tangents drawn from an external point to a circle are equal]

⇒ QB = QC

⇒ PQ bisects BC.

Question 7

In figure, tangents PQ and PR are drawn to a circle such that ∠RPQ = 30°. A chord RS is drawn parallel to the tangent PQ. Find the ∠RQS.

Solution:

PQ and PR are two tangents drawn from an external point P.

∴ PQ = PR

[Lengths of tangents drawn from an external point to a circle are equal]

⇒ ∠PQR = ∠QRP

[Angles opposite to equal sides are equal]

Now, in ∆PQR,

∠PQR + ∠QRP + ∠RPQ = 180° [Angle sum property]

⇒ ∠PQR + ∠PQR + 30° = 180°

⇒ 2∠PQR = 180° – 30° = 150°

[∵ ∠PQR = ∠QRP]

⇒ ∠PQR = 75°

Since, SR ॥ QP

∠SRQ = ∠RQP = 75° [Alternate interior angles]

Also, ∠PQR = ∠QSR = 75° [Angles in alternate segment]

In ∆QRS,

∠Q + ∠R + ∠S = 180° [Angle sum property]

⇒ ∠Q = 180° – (75° + 75°) = 30°

∴ ∠RQS = 30°

Question 8

AB is a diameter and AC is a chord of a circle with centre O such that ∠BAC = 30°.The tangent at C intersects extended AB at a point D. Prove that BC = BD.

Solution:

Given, AB is a diameter and AC is a chord of circle with centre O, ∠BAC = 30°

To prove : BC = BD

Join BC

∠BCD = ∠CAB [Angles in alternate segment]

∠CAB = 30° [Given]

∠BCD = 30° …(i)

∠ACB = 90° [Angle in semi-circle]

In ∆ABC,

∠A + ∠B + ∠C = 180° [Angle sum property]

30° + ∠CBA + 90° = 180°

⇒ ∠CBA = 60°

Also, ∠CBA + ∠CBD = 180° [Linear pair]

⇒ ∠CBD = 180° – 60° = 120°

[∵ ∠ CBA = 60°]

Now, in ACBD,

∠CBD + ∠BDC + ∠DCB = 180°

⇒ 120° + ∠BDC + 30° = 180°

⇒ ∠BDC = 30° …(ii)

From (i) and (ii),

∠BCD = ∠BDC

⇒ BC = BD

[Sides opposite to equal angles are equal]

Question 9

Prove that the tangent drawn at the mid-point of an arc of a circle is parallel to the chord joining the end points of the arc.

Solution:

Let mid-point of an arc AMB be M and TMT’ be the tangent to the circle.

Join AB, AM and MB.

Since arc AM = arc MB =3 Chord AM = Chord MB

In ∆AMB, AM = MB

⇒ ∠MAB = ∠MBA …(i)

[Sides opposite to equal angles are equal]

Since, TMT’ is a tangent line.

∴ ∠AMT = ∠MBA [Angles in alternate segments are equal]

= ∠MAB [from equation (i)]

But ∠AMT and ∠MAB are alternate angles, which is possible only when AB ॥ TMT

Hence, the tangent drawn at the mid-point of an arc of a circle is parallel to the chord joining the end points of the arc.

Question 10

In figure, the common tangent, AB and CD to two circles with centres O and O’ intersect at E. Prove that the points O, E, O’ are collinear.

Solution:

Join AO, OC and O’D, O’B.

Now, in ∆EO’D and ∆EO’B,

O’D = O’B

O’E = O’E

ED = EB

[Tangents drawn from an external point to the circle are equal in length]

∴ EO’D ≅ ∆ EO’B [By SSS congruence criterion]

⇒ ∠O’ED = ∠O’EB …(i)

i.e., O’E is the angle bisector of ∠DEB.

Similarly, OE is the angle bisector of ∠AEC. Now, in quadrilateral DEBO’.

∠O’DE = ∠O’BE = 90°

[CED is a tangent to the circle and O’D is the radius, i.e., O’D ⊥ CED]

⇒ ∠O’DE + ∠O’BE = 180°

∴ ∠DEB + ∠DO’B = 180°

[∵ DEBO’ is cyclic quadrilateral] …(ii)

Since, AB is a straight line.

∴ ∠AED + ∠DEB = 180°

⇒ ∠AED + 180° – ∠DO’B = 180° [from (ii)]

⇒ ∠AED = ∠DO’B …(iii)

Similarly, ∠AED = ∠AOC …(iv)

Again from eq. (ii), ∠DEB = 180° – ∠DO’B

Dividing by 2 on both sides, we get

So, OEO’ is straight line.

Hence, O, E and O’ are collinear.

Question 11

In figure, O is the centre of a circle of radius 5 cm, T is a point such that 07= 13 cm and 07 intersects the circle at E. If AB is the tangent to the circle at E, find the length of AB.

Solution:

OP is perpendicular to PT.

∴ In ∆OPT,

OT^{2} = OP^{2} + PT^{2}

⇒ PT^{2} = OT^{2} – OP^{2}

⇒ PT^{2} = (13)^{2} – (5)^{2} = 169 – 25 = 144

⇒ PT= 12 cm

Since, the length of pair of tangents from an external point T is equal.

∴ QT= 12 cm

Now, TA = PT – PA

⇒ TA = 12 -PA …(i)

and TB = QT – QB

TB = 12 – QB …(ii)

Also, PA = AE and QB = EB …(iii) [Pair of tangents]

∴ ET = OT – OE [∵ OE = 5 cm = radius]

⇒ ET = 13 – 5

⇒ ET = 8 cm

Since, AB is a tangent and OE is the radius.

∴OE ⊥ AB

⇒ ∠OEA = 90°

∴ ∠AET = 180° – ∠OEA [Linear pair]

⇒ ∠AET = 90°

Now, in right angled ∆AET,

(AT)^{2} = (AE)^{2} + (ET)^{2} [by Pythagoras theorem]

⇒ (12 – PA)^{2} = (PA)^{2} + (8)^{2}

⇒ 144 + (PA)^{2} – 24 PA = (PA)^{2} + 64

⇒ 24 PA = 80

Question 12

The tangent at a point C of a circle and a diameter AB when extended intersect at P. If ∠PCA = 110°, Find ∠CBA.

Solution:

Join OC. Here, OcC is the redius.

Since, tangent at any point of a circle is – perpendicular to the radius through the point of contact.

∴ OC ⊥ PC

Now, ∠PCA = 110° [Given]

⇒ ∠PCO + ∠OCA = 110°

⇒ 90° + ∠OCA = 110°

⇒ ∠OCA = 20°

∵ OC = OA = radius of circle

∴ ∠OCA = ∠OAC = 20°

[Sides opposite to equal angles are equal]

Since, PC is a tangent,

∠BCP = ∠CAB = 20° [Angles in alternate segment]

In ∆PAC,

∠P + ∠C + ∠A = 180°

∠P = 180° – (∠C + ∠A)

= 180°-(110°+ 20°)

= 180° – 130° = 50°

In ∆PBC,

∠BPC + ∠PCB + ∠CBP = 180°

⇒ 50° + 20° + ∠PBC = 180°

⇒ ∠PBC = 180° – 70°

⇒ ∠PBC = 110°

Since, ABP is a straight line.

∴ ∠PBC + ∠CBA = 180°

⇒ ∠CBA = 180° – 110° = 70°

Question 13

if an isosceles triangle ABC, in which AB = AC = 6 cm, is inscribed in a circle of radius 9 cm, find the area of the triangle.

Solution:

Join OB, OC and OA.

In ∆ABO and ∆ACO,

AB = AC [Given]

BO = CO [Radii of same circle]

AO = AO [Common side]

∴ ∆ABO ≅ ∆ACO [By SSS congruence criterion]

⇒ ∠1 = ∠2 [CPCT]

Now, in ∆ABM and ∆ACM,

AB = AC [Given]

∠1 = ∠2 [proved above]

AM = AM [Common side]

∴ ∆AMB ≅ ∆AMC [By SAS congruence criterion]

⇒ ∠AMB = ∠AMC [CPCT]

Also, ∠AMB + ∠AMC = 180° [Linear pair]

⇒ ∠AMB = 90°

We know that a perpendicular from the centre of circle bisects the chord.

So, OA is a perpendicular bisector of BC.

Let AM = x, then OM = 9 – x [ ∵ OA = radius = 9 cm]

In right angledd ∆AMC,

AC^{2} = AM^{2} + MC^{2}

[By Pythagoras theorem]

⇒ MC^{2} = 6^{2} – x^{2} …(i)

In right angle ∆OMC,

OC^{2} = OM^{2} + MC^{2} [By Pythagoras theorem]

⇒ MC^{2} = 9^{2} – (9 – x)^{2}

From equation (i) and (ii),

6^{2} – x^{2} = 9^{2} – (9 – x)^{2}

⇒ 36 – x^{2} = 81 – (81 + x^{2} – 18x)

⇒ 36 = 18x

⇒ x = 2

∴ AM = 2 cm

From equation (ii),

MC^{2} = 9^{2} – (9 – 2)^{2}

⇒ MC^{2} = 81 – 49 = 32

⇒ MC = 4√2 cm

∴ BC = 2 MC = 8√2 cm

Fience, the required area of ∆ABC is 8√2 cm^{2}.

Question 14

1 is a point at a distance 13 cm from the centre 0 of a circle of radius 5 cm. AP and AQ are the tangents to the circle at P and Q. If a tangent BC is drawn at a point R lying on the minor arc PQ to intersect AP at B and AQ at C, find the perimeter of the ∆ABC.

Solution:

OP ⊥ AP

∴ ∠OPA = 90°

[Tangent at any point of a circle is perpendicular to the radius through the point of contact]

In ∆OAP,

OA^{2} = OP^{2} + PA^{2}

⇒ 13^{2} = 5^{2} + PA^{2}

⇒ PA = 12 cm

Now, perimeter of ∆ABC = AB + BC + CA

= AB + BR + RC + CA

= (AB + BR) + (RC + CA)

= (AB + BP) + (CQ + CA)

[ ∵ BR = BP, RC = CQ i.e., tangents from external point to a circle are equal]

= AP + AQ

= 2AP [∵ AP = AQ]

= 2 x 12 = 24 cm

Hence, the perimeter of ∆ABC = 24 cm.

We hope the NCERT Exemplar Class 10 Maths Chapter 9 Circles will help you. If you have any query regarding NCERT Exemplar Class 10 Maths Chapter 9 Circles, drop a comment below and we will get back to you at the earliest.